How many triangles with positive area have all their vertices at points $(i,j)$ in the coordinate plane, where $i$ and $j$ are integers between $1$ and $5$, inclusive?
$\textbf{(A)}\ 2128 \qquad\textbf{(B)}\ 2148 \qquad\textbf{(C)}\ 2160 \qquad\textbf{(D)}\ 2200 \qquad\textbf{(E)}\ 2300$

Answer: We can solve this by finding all the combinations, then subtracting the ones that are on the same line. There are $25$ points in all, from $(1,1)$ to $(5,5)$, so $\dbinom{25}3$ is $\frac{25\cdot 24\cdot 23}{3\cdot 2 \cdot 1}$, which simplifies to $2300$. Now we count the ones that are on the same line. We see that any three points chosen from $(1,1)$ and $(1,5)$ would be on the same line, so $\dbinom53$ is $10$, and there are $5$ rows, $5$ columns, and $2$ long diagonals, so that results in $120$. We can also count the ones with $4$ on a diagonal. That is $\dbinom43$, which is 4, and there are $4$ of those diagonals, so that results in $16$. We can count the ones with only $3$ on a diagonal, and there are $4$ diagonals like that, so that results in $4$. We can also count the ones with a slope of $\frac12$, $2$, $-\frac12$, or $-2$, with $3$ points in each. There are $12$ of them, so that results in $12$. Finally, we subtract all the ones in a line from $2300$, so we have $2300-120-16-4-12=\boxed{2148}$.